No, -8 - 2(3 + 2n) + 7n is not equivalent to -30 - 13n
Step-by-step explanation:
Let us revise the operation of the negative and positive numbers
- (-) + (-) = (-)
- (-) × (-) = (+)
- (-) + (+) = the sign of greatest [(-) if the greatest is (-) or (+) if the greatest is (+)]
- (-) × (+) = (-)
- (-) - (+) = (-)
- (+) - (-) = (+)
∵ The expression is -8 - 2(3 + 2n) + 7n
- Simplify it
∵ 2(3 + 2n) = 2(3) + 2(2n) = 6 + 4n
∴ -8 - 2(3 + 2n) + 7n = -8 - (6 + 4n) + 7n
- Multiply the bracket by (-)
∴ -8 - (6 + 4n) + 7n = -8 - 6 - 4n + 7n
- Add the like terms
∴ -8 - (6 + 4n) + 7n = (-8 - 6) - 4n + 7n
∴ -8 - (6 + 4n) + 7n = -14 + 3n
∴ -8 - 2(3 + 2n) + 7n is equivalent to -14 + 3n
∵ -14 + 3n ≠ -30 - 13n
∴ -8 - 2(3 + 2n) + 7n is not equivalent to -30 - 13n
No, -8 - 2(3 + 2n) + 7n is not equivalent to -30 - 13n
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You can learn more about the directed numbers in brainly.com/question/10364988
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Answer:
The best point of estimate for the true mean is:
Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.
Step-by-step explanation:
Information given
represent the sample mean for the late time for a flight
population mean
represent the population deviation
n=76 represent the sample size
Confidence interval
The best point of estimate for the true mean is:
The confidence interval for the true mean is given by:
(1)
The Confidence level given is 0.95 or 95%, th significance would be 
and 
. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got 
Replacing we got:
Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.
Answer:
There is no sufficient evidence to reject the company's claim at the significance level of 0.05
Step-by-step explanation:
Let be the true mean weight per apple the company ship. We want to test the next hypothesis
vs 
(two-tailed test).
Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by
which is normally distributed. The observed value is
. The rejection region for 
is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.
Answer:
Hey there!
You can think of the rate of change as the slope of a quadratic function- here we see that it is 9/-3, or - 3.
Let me know if this helps :)
